From dd4b8ebf78239e4743ccefe69a86d64b42887e4f Mon Sep 17 00:00:00 2001 From: Meiling WU Date: Mon, 30 May 2022 09:33:27 +0000 Subject: [PATCH] Update toy_document_en.Rmd --- module2/exo1/toy_document_en.Rmd | 47 ++++++++------------------------ 1 file changed, 12 insertions(+), 35 deletions(-) diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd index 8108e81..bae961c 100644 --- a/module2/exo1/toy_document_en.Rmd +++ b/module2/exo1/toy_document_en.Rmd @@ -1,50 +1,27 @@ --- -title: "On the computation of pi" -Author: Pedro Mota -Date: 14.01.2021 -output: - html_document: - df_print: paged +title: "How" +Author: Meiling WU +Date: 30.05.2022 +output: html_document: --- -## Asking the math library -My computer me that π is *approximatively* - -```{r} -pi +```{r setup, include=FALSE} +knitr::opts_chunk$set(echo = TRUE) ``` -## Buffon's needle - -Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem) we get the **aproximation** - -```{r} -set.seed(42) -N = 100000 -x = runif(N) -theta = pi/2*runif(N) -2/(mean(x+sin(theta)>1)) -``` +## Some explanations -## Using a surface fraction argument -A method that is easier to understand and does not make use of the **sin** function is based on the fact that if $X∼U(0,1)$ and $Y∼U(0,1)$, then $P[X2+Y2≤1]=π/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The Following code use this approach: -```{r} -set.seed(42) -N = 1000 -df = data.frame(X = runif(N), Y = runif(N)) -df$Accept = (df$X**2 + df$Y**2 <=1) -library(ggplot2) -ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw() -``` -It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, $X2+Y2$ is smaller than 1 : -```{r} -4*mean(df$Accept) +```{r cars} +summary(cars) ``` +```{r pressure, echo=FALSE} +plot(pressure) +``` -- 2.18.1